Chapter 2 — Relations and Functions

Concept Theory

A relation from set \(A\) to set \(B\) is a subset of the Cartesian product \(A\times B\). Each element of a relation is an ordered pair.

For a relation \(R\):

• Domain → set of all first elements of ordered pairs.
• Codomain → target set where relation is defined.
• Range → set of actual second elements obtained.

If \(R\subseteq A\times A\), then:

Domain ⊆ A, Range ⊆ A, Codomain = A

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Solution Roadmap

• Rewrite the relation equation.
• Find values of \(x\) such that \(y\) remains inside set \(A\).
• Construct valid ordered pairs.
• Extract Domain and Range from those pairs.

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Solution

Given

\[ A=\{1,2,3,\ldots,14\} \]

Relation:

\[ R=\{(x,y):3x-y=0,\ x,y\in A\} \]

From the condition

\[ 3x-y=0 \]

\[ y=3x \]

Thus for every value of \(x\), the corresponding value of \(y\) must be \(3x\). However, \(y\) must also belong to set \(A\).

So we require

\[ 3x \le 14 \]

Possible values of \(x\):

\[ x=1,2,3,4 \]

Corresponding values of \(y\):

\[ 3,6,9,12 \]

Hence the relation is

\[ R=\{(1,3),(2,6),(3,9),(4,12)\} \]

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Domain

\[ \{1,2,3,4\} \]

Range

\[ \{3,6,9,12\} \]

Codomain

\[ \{1,2,3,\ldots,14\} \]

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Relation Mapping Illustration

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Exam Significance

Understanding domain and range of relations is essential for:

• CBSE Board Exams – direct questions on relations and functions.
• JEE Main / CUET – conceptual questions on domain, range and mapping.
• NDA / SSC – identification of ordered pairs and relation sets.

This concept forms the foundation for later topics such as:

• Functions
• Inverse functions
• Mapping diagrams
• Cartesian products

Concept Theory

A relation on a set of natural numbers is a collection of ordered pairs that satisfy a given rule. If a relation is defined using an equation or condition, we convert it into roster form by listing all ordered pairs that satisfy the rule.

For a relation \(R\):

• Domain → set of all first elements of ordered pairs.
• Range → set of all second elements of ordered pairs.
• Roster Form → representation of relation by explicitly listing ordered pairs.

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Solution Roadmap

• Identify all valid values of \(x\).
• Substitute those values in the relation equation \(y=x+5\).
• Construct ordered pairs.
• Write the relation in roster form.
• Extract domain and range.

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Solution

The relation is defined as

\[ R=\{(x,y):y=x+5,\ x\in\mathbb{N},\ x<4\} \]

Since \(x\) is a natural number less than \(4\), the possible values of \(x\) are

\[ x=1,2,3 \]

Substituting these values into \(y=x+5\):

\[ \begin{aligned} x=1 &\Rightarrow y=6 \ x=2 &\Rightarrow y=7 \ x=3 &\Rightarrow y=8 \end{aligned} \]

Therefore the relation in roster form is

\[ R=\{(1,6),(2,7),(3,8)\} \]

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Domain

\[ \{1,2,3\} \]

Range

\[ \{6,7,8\} \]

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Relation Mapping Illustration

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Exam Significance

This type of problem tests the fundamental understanding of relations and ordered pairs.

• CBSE Board Exams – direct questions on roster form and domain–range.
• JEE Main / CUET – problems involving relations defined by equations.
• NDA / SSC – conceptual questions on ordered pairs and mappings.

Mastery of these concepts is essential before studying:

• Functions
• One–one and onto mappings
• Inverse functions
• Cartesian products

h4 style="color:#7dd3fc;">Concept Theory

The difference between two integers is odd when one number is even and the other is odd.

Thus,

• odd − even = odd
• even − odd = odd

If both numbers are even or both are odd, their difference will be even.

A relation from set \(A\) to set \(B\) consists of ordered pairs \((x,y)\) where \(x\in A\) and \(y\in B\) satisfying the given condition.

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Solution Roadmap

• Identify odd and even numbers in sets \(A\) and \(B\).
• Form ordered pairs \((x,y)\) from \(A\times B\).
• Select pairs where the difference between numbers is odd.
• List these pairs in roster form.

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Solution

Given sets

\[ A=\{1,2,3,5\} \]

\[ B=\{4,6,9\} \]

Identify parity of elements:

• Odd numbers in \(A\): 1,3,5
• Even numbers in \(A\): 2
• Even numbers in \(B\): 4,6
• Odd numbers in \(B\): 9

Now we form ordered pairs where one number is even and the other is odd.

Checking each possible pair:

\[ \begin{aligned} (1,4), (1,6) \ (2,9) \ (3,4), (3,6) \ (5,4), (5,6) \end{aligned} \]

Hence the relation in roster form is

\[ R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\} \]

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Parity Mapping Illustration

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Exam Significance

Problems involving relations defined by numerical conditions test the ability to analyse properties such as parity (odd/even), divisibility, or inequalities.

• CBSE Board Exams – direct questions on writing relations in roster form.
• JEE Main / CUET – identifying ordered pairs satisfying given constraints.
• NDA / SSC – logic based problems involving number properties.

This concept is foundational for later topics such as:

• Cartesian products
• Relations and functions
• One-one and many-one mappings

Concept Theory

A relation between two sets can be represented using a mapping diagram. In such diagrams, arrows show how elements of one set are connected to elements of another set.

• Roster form lists all ordered pairs explicitly.
• Set-builder form describes the relation using a rule or formula.
• Domain → set of first elements.
• Range → set of second elements actually mapped.

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Solution Roadmap

• Observe the mapping arrows between sets \(P\) and \(Q\).
• Write the ordered pairs shown in the diagram.
• Identify the rule connecting elements.
• Write the relation in roster and set-builder form.
• Extract domain and range.

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Exam Significance

Mapping-diagram questions test the ability to interpret relations visually.

• CBSE Board Exams frequently ask students to convert mapping diagrams into roster form.
• JEE Main / CUET questions may require identifying the rule of a relation from ordered pairs.
• Foundation mathematics – this concept leads to functions and inverse relations.

Understanding such diagrams helps students analyse relations, mappings, and functional rules quickly in competitive examinations.

A relation defined by divisibility means that for ordered pair \((a,b)\), the number \(b\) must be a multiple of \(a\).

Mathematically, \(b\) is divisible by \(a\) if

\[ b = a \times k \]

for some integer \(k\).

• Roster form lists all valid ordered pairs.
• Domain → set of first elements.
• Range → set of second elements.

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Solution Roadmap

• Take each element of set \(A\) as \(a\).
• Find elements \(b\in A\) divisible by \(a\).
• Construct ordered pairs \((a,b)\).
• Write the relation in roster form.
• Extract domain and range.

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Solution

Given set

\[ A=\{1,2,3,4,6\} \]

The relation is defined as

\[ R=\{(a,b):a,b\in A,\ b \text{ is divisible by } a\} \]

Now we check divisibility for each element of \(A\).

Multiples within the set:

• Multiples of 1 → 1,2,3,4,6
• Multiples of 2 → 2,4,6
• Multiples of 3 → 3,6
• Multiples of 4 → 4
• Multiples of 6 → 6

Thus the relation in roster form is

\[ R=\{(1,1),(1,2),(1,3),(1,4),(1,6), (2,2),(2,4),(2,6), (3,3),(3,6), (4,4), (6,6)\} \]

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Domain

\[ \{1,2,3,4,6\} \]

Range

\[ \{1,2,3,4,6\} \]

Thus the domain and range are equal to the set \(A\).

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Divisibility Relation Visualization

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Exam Significance

Relations based on divisibility are frequently used to test logical reasoning in mathematics.

• CBSE Board Exams – direct questions on writing relations in roster form.
• JEE Main / CUET – divisibility relations appear in sets and number theory questions.
• Olympiad / NDA – relations defined by number properties.

Understanding divisibility relations also helps in studying:

• Partial ordering relations
• Equivalence relations
• Number theory concepts

Concept Theory

Relations are often defined using algebraic rules. If a relation is expressed in the form

\[ R=\{(x,f(x))\} \]

then:

• Domain → all possible values of \(x\).
• Range → values obtained after applying the rule \(f(x)\).

In this problem the rule is

\[ y=x+5 \]

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Solution Roadmap

• Identify all values of \(x\).
• Substitute each value into \(x+5\).
• Construct ordered pairs.
• Extract domain and range.

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Solution

Given relation

\[ R=\{(x,x+5):x\in\{0,1,2,3,4,5\}\} \]

Substituting each value of \(x\):

\[ \begin{aligned} (0,5)\ (1,6)\ (2,7)\ (3,8)\ (4,9)\ (5,10) \end{aligned} \]

Thus the relation in roster form is

\[ R=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\} \]

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Domain

\[ \{0,1,2,3,4,5\} \]

Range

\[ \{5,6,7,8,9,10\} \]

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Relation Mapping Illustration

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Exam Significance

Relations defined by algebraic expressions are important because they form the foundation of the concept of functions.

• CBSE Board Exams – identifying domain and range from algebraic relations.
• JEE Main / CUET – evaluating ordered pairs using function-like rules.
• Foundation mathematics – understanding mappings and functions.

Such problems prepare students for later topics like:

• Functions and graphs
• Inverse functions
• Composite functions

Concept Theory

A relation defined by a rule such as \(R=\{(x,f(x))\}\) requires evaluating the function \(f(x)\) for all permissible values of \(x\).

Here the rule is the cubic function:

\[ y=x^3 \]

Only values of \(x\) that satisfy the condition “prime number less than 10” are used.

• A prime number has exactly two positive divisors: 1 and itself.
• The primes less than 10 are: 2, 3, 5, 7.

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Solution Roadmap

• Identify all prime numbers less than 10.
• Substitute each value of \(x\) into \(x^3\).
• Construct ordered pairs.
• Write the relation in roster form.

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Solution

The relation is defined as

\[ R=\{(x,x^3):x\text{ is a prime number less than }10\} \]

Prime numbers less than 10 are

\[ 2,3,5,7 \]

Evaluating the cube of each value:

\[ \begin{aligned} 2^3 &= 8\ 3^3 &= 27\ 5^3 &= 125\ 7^3 &= 343 \end{aligned} \]

Thus the relation in roster form is

\[ R=\{(2,8),(3,27),(5,125),(7,343)\} \]

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Mapping Illustration

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Exam Significance

This question combines two mathematical ideas: prime numbers and relations defined by algebraic rules.

• CBSE Board Exams – identifying elements satisfying given conditions.
• JEE Main / CUET – combining number theory with functions or relations.
• Foundation mathematics – understanding how functions generate relations.

Such problems help build intuition for topics like:

• Functions and graphs
• Polynomial mappings
• Number theory patterns

Concept Theory

A relation from set \(A\) to set \(B\) is defined as any subset of the Cartesian product \(A\times B\).

If

\[ n(A)=m,\quad n(B)=n \]

then the Cartesian product contains

\[ n(A\times B)=m\times n \]

elements.

Since every relation is a subset of \(A\times B\), the total number of relations is equal to the number of subsets of \(A\times B\).

\[ \text{Number of relations}=2^{\,n(A\times B)} \]

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Solution Roadmap

• Find the number of elements in sets \(A\) and \(B\).
• Compute the size of the Cartesian product \(A\times B\).
• Use the subset formula to determine the number of relations.

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Solution

Given sets

\[ A=\{x,y,z\}, \quad B=\{1,2\} \]

Number of elements:

\[ n(A)=3, \quad n(B)=2 \]

The Cartesian product contains

\[ n(A\times B)=n(A)\times n(B) \]

\[ =3\times 2 \]

\[ =6 \]

Thus \(A\times B\) contains 6 ordered pairs.

A relation from \(A\) to \(B\) is any subset of \(A\times B\).

\[ \text{Number of relations}=2^{6} \]

\[ =64 \]

Hence, the number of relations from \(A\) to \(B\) is

\[ \boxed{64} \]

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Cartesian Product Visualization

The diagram illustrates the 6 ordered pairs in \(A\times B\).

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Exam Significance

Counting the number of relations is an important combinatorial concept in set theory.

• CBSE Board Exams frequently test the formula \(2^{mn}\).
• JEE Main / CUET problems often combine relations with combinatorics.
• Foundation mathematics – important for functions, relations, and discrete mathematics.

Understanding this idea also helps in studying:

• Functions and mappings
• Binary relations
• Discrete mathematics and graph theory

Concept Theory

The set of integers \(\mathbb{Z}\) has an important property called closure under subtraction.

This means:

\[ a,b \in \mathbb{Z} \Rightarrow a-b \in \mathbb{Z} \]

Therefore, the difference of any two integers is always an integer.

When a relation condition is always satisfied, the relation contains all possible ordered pairs of the given sets.

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Solution Roadmap

• Analyse the condition \(a-b\) is an integer.
• Recall the closure property of integers under subtraction.
• Conclude that every pair of integers satisfies the condition.
• Determine domain and range.

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Solution

The relation is defined as

\[ R=\{(a,b):a,b\in\mathbb{Z},\ a-b \text{ is an integer}\} \]

Since integers are closed under subtraction,

\[ a-b \in \mathbb{Z} \]

for every pair \(a,b\in\mathbb{Z}\).

Thus the condition is satisfied for all ordered pairs of integers.

\[ R=\mathbb{Z}\times\mathbb{Z} \]

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Domain

\[ \mathbb{Z} \]

Range

\[ \mathbb{Z} \]

Thus both the domain and range of the relation are the set of all integers.

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Relation Visualization

Every integer can relate to every other integer because subtraction remains an integer.

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Exam Significance

This question highlights the closure property of integers, which is an important concept in algebra and set theory.

• CBSE Board Exams – conceptual understanding of relations on number sets.
• JEE Main / CUET – recognising when a relation becomes the full Cartesian product.
• Foundation mathematics – understanding properties of integers and algebraic structures.

Such concepts are later used in:

• equivalence relations
• algebraic structures
• abstract algebra